Consider the following reactionN2( g)+3H2( g)→2NH3( g). The rate of thereaction in terms of N2 at T(k) is -dN2dt=0.02 mole. lit-1·sec-1. What is the value of -dH2dt (in mole. lit-1·sec-1 ) at the same temperature?

N2(g)+3H2(g)⇔2NH3(g) −d[N2]dt=−13d[H2]dt −d[H2]dt=3(−d[N2]dt) =3×0.02 =0.06 −d[H2]dt=0.06

Consider the following reactionN2( g)+3H2( g)→2NH3( g). The rate of thereaction in terms of N2 at T(k) is -dN2dt=0.02 mole. lit-1·sec-1. What is the value of -dH2dt (in mole. lit-1·sec-1 ) at the same temperature?

N2(g)+3H2(g)⇔2NH3(g) −d[N2]dt=−13d[H2]dt −d[H2]dt=3(−d[N2]dt) =3×0.02 =0.06 −d[H2]dt=0.06

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Consider the following reaction
$N_{2 (g)} + 3 H_{2 (g)} \to 2 {NH}_{3 (g)} .$
The rate of thereaction in terms of $N_{2}$ at $T (k)$ is $- \frac{d [N_{2}]}{dt} = 0.02$ mole. ${lit}^{- 1} \cdot \sec^{- 1}$ . What is the value of $- \frac{d [H_{2}]}{dt}$ (in mole. ${lit}^{- 1} \cdot \sec^{- 1}$ ) at the same temperature?

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0.02

0.06

0.04

answer is C.

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Detailed Solution

$\begin{aligned} N_{2} (g) + 3 H_{2} (g) \Leftrightarrow 2 {NH}_{3} (g) \\ \frac{- d [N_{2}]}{dt} = - \frac{1}{3} \frac{d [H_{2}]}{dt} \\ \frac{- d [H_{2}]}{dt} = 3 (\frac{- d [N_{2}]}{dt}) \\ = 3 \times 0.02 \\ = 0.06 \\ \frac{- d [H_{2}]}{dt} = 0.06 \end{aligned}$